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The x intercepts are the roots of the polynomial f(x) = 0.
The fundamental theorem of algebra says that the number of roots of this equation is 9, but not necessarily all of them are real; therefore the MAXIMUM number of x intercepts is 9.
However, since f(x) is of odd degree it must have at least one real root; therefore, the MINIMUM number of x intercepts is 1.
I believe the answer to be "107"
Answer:
sin2x + 
x + c
Step-by-step explanation:
using the trigonometric identity
• cos² x = (cos2x + 1) and
∫cos ax = 
sin ax
hence
∫(1 - (cos2x + 1)dx
= ∫(1 - cos2x - )dx
= ∫( - cos2x)dx
= x - sin2x + c
where c is the constant of integration
4/5 x 3/4 is just 12/20, and then when you simplify that you get 3/4.
2c - 5 < -21 - 2c |add 5 to both sides
2c < -16 - 2c |add 2c to both sides
4c < -16 |divide both sides by 4
c < -4
